There's a standard trick using the Yoneda lemma for computing what universal objects in (restricted) functor categories must be if they exist. In the case of the subobject classifier, this is explained on p. 37 of Sheaves in Geometry and Logic.
Specifically, suppose that $\Omega\colon \mathbf{N}\to \mathbf{FinSets}$ is a subobject classifier in $\mathbf{FinSets}^\mathbf{N}$. Let's try to find out what finite set $\Omega(0)$ is. Since $\mathbf{FinSets}^\mathbf{N}$ is a full subcategory of $\mathbf{Sets}^\mathbf{N}$, by Yoneda we have $$\Omega(0) \cong \text{Hom}_{\mathbf{Sets}^\mathbf{N}}(h^0,\Omega) = \text{Hom}_{\mathbf{FinSets}^\mathbf{N}}(h^0,\Omega) \cong \text{Sub}(h^0),$$ where $h^0$ is the functor $\text{Hom}_{\mathbf{N}}(0,-)$. But now $h^0$ has infinitely many subobjects, but $\Omega(0)$ is a finite set, and this is a contradiction.
To see that $h^0$ has infinitely many subobjects, just note that since $0$ is the initial object in $\mathbf{N}$, $h^0(n)$ is a singleton $\{*\}$ for all $n$. Incidentally, this makes $h^0$ isomorphic to the terminal object $1$ - your guess about the identity of terminal object is correct.
Now for each natural number $n$ (or $n = \infty$), there is a distinct subobject of $h^0$, given by $$m\mapsto \begin{cases} \varnothing & \text{if }m<n\\ \{*\} & \text{if }m\geq n.\end{cases}$$
First of all, when the hint speaks of a "quasi-inverse" it is referring to the following equivalent of the given definition: a functor $F : \mathbf{C} \to \mathbf{D}$ is an equivalence of categories if and only if there exists a functor $G : \mathbf{D} \to \mathbf{C}$ such that $F \circ G \simeq \operatorname{id}_{\mathbf{D}}$ and $G \circ F \simeq \operatorname{id}_{\mathbf{C}}$; and in this case, $G$ is called a quasi-inverse of $F$.
So, one way to follow the hint would be to explain how $\mathbf{s}$ becomes a functor (i.e. how it operates on morphisms, and show it preserves identities and compositions), and then establish isomorphisms $\mathbf{r} \circ \mathbf{s} \simeq \operatorname{id}$ and $\mathbf{s} \circ \mathbf{r} \simeq \operatorname{id}$.
On the other hand, it is possible to proceed using the definition you stated. First, as a preliminary, I don't know if MacLane and Moerdik specified what exactly $\operatorname{Sh}(\mathcal{B})$ means; but the reasonable definition would be that it is the presheaves on the poset category of $\mathcal{B}$ such that whenever $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$ is a cover of $U \in \mathcal{B}$, we have an equalizer diagram
$$F(U) \rightarrow \prod_{i\in I} F(V_i) \rightrightarrows \prod_{i, j \in I, W\in \mathcal{B}, W \subseteq V_i \cap V_j} F(W).$$
(The first step would be to see why $\mathbf{r}$ of a sheaf on $X$ would satisfy this condition; I will leave that as an exercise.)
So, first let us see that $\mathbf{r}$ is injective on morphisms; so, suppose that we have two morphisms $f, g : F \to G$ such that $f(V) = g(V)$ whenever $V \in \mathcal{B}$. Then for any open $U$ and $x \in F(U)$, there is a cover of $U$ by elements $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. Now, by the hypothesis, $$f(x) {\mid_{V_i}} = f(V_i)(x {\mid_{V_i}}) = g(V_i)(x {\mid_{V_i}}) = g(x) {\mid_{V_i}}$$ for each $i$; and by the injectivity part of the equalizer condition defining that $G$ is a sheaf, we conclude that $f(x) = g(x)$. Since this is true for any open $U$ and any $x \in F(U)$, then $f = g$.
Similarly, to see that $\mathbf{r}$ is surjective on morphisms, suppose we have $f : \mathbf{r}(F) \to \mathbf{r}(G)$. Then for any open $U \subseteq X$ and $x \in F(U)$, again choose a cover of $U$ by $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. (In fact, to forestall questions of the following construction being well-defined, let us use the canonical maximal cover of all elements of $\mathcal{B}$ contained in $U$.) Then for each $i \in I$, define $y_i := f(V_i)(x {\mid_{V_i}})$. Then for each $i,j$, we can find the canonical maximal cover of $V_i \cap V_j$ by $\{ W_k \mid k \in K_{i,j} \} \subseteq \mathcal{B}$. Now for each $k$, we have
$$y_i {\mid_{W_k}} = f(V_i)(x {\mid_{V_i}}) {\mid_{W_k}} = f(W_k)((x {\mid_{V_i}}) {\mid_{W_k}}) = F(W_k)(x {\mid_{W_k}}) = y_j {\mid_{W_k}}.$$
Therefore, by the injectivity part of the sheaf condition on $G$, we have $y_i {\mid_{U_i \cap U_j}} = y_j {\mid_{U_i \cap U_j}}$. Then, by the exactness part of the sheaf condition on $G$, there exists a unique $y \in G(U)$ such that $y {\mid_{U_i}} = y_i$. We now define $f'(U)(x) := y$.
It remains to show that $f'$ defines a morphism of sheaves, and that $\mathbf{r}(f') = f$. (Hint for the morphism of sheaves part: given $U' \subseteq U$ and $x \in F(U)$, show that $(f'(U) {\mid_{U'}}) {\mid_{V_i}}$ is equal to $y_i$ when you put $x {\mid_{U'}}$ in place of $x$, and then apply the injectivity part of the sheaf condition on $G$.)
Now, to show that $\mathbf{r}$ is essentially surjective, suppose we have $F \in \operatorname{Sh}(\mathcal{B})$. Then for each open $U$, define $G(U)$ to be the equalizer in the diagram
$$G(U) \rightarrow \prod_{V \in \mathcal{B}, V \subseteq U} F(V) \rightrightarrows \prod_{V, V', W \in \mathcal{B}, V \subseteq U, V' \subseteq U, W \subseteq V \cap V'} F(W).$$
The restriction maps of $G$ will then be constructed based on the universal property of equalizers. We now need to see that $G$ is a sheaf on $X$, and that $\mathbf{r}(G) \simeq F$. The latter follows fairly directly from the sheaf condition on $F$.
For the sheaf condition, suppose we have a cover $\{ U_i \mid i \in I \}$ of $U$ and sections $x_i \in G(U_i)$ such that $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ for each $i,j$. Then each $x_i$ can be decomposed into the compatible data of an element of $F(V)$ for every $V \in \mathcal{B}$, $V \subseteq U_i$ which we will call $x_i {\mid_V}$. But then, the union of the canonical covers of each $U_i$ will form a cover of $U$; and for each $W$ in this cover, we can choose $i$ such that $W \subseteq U_i$, and define $y_W := x_i {\mid_W}$. If we have two different indices $i,j$ such that $W \subseteq U_i$ and $W \subseteq U_j$, then from the condition $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ we get $x_i {\mid_W} = x_j {\mid_W}$, which makes this definition of $y_V$ well-defined. Once we verify the compatibility condition on $(y_W)$, we get a section $z_V \in F(V)$ from the definition of $F$ being a sheaf. It now remains to show that this family of $z_V$ satisfies the compatibility condition from the definition of $G$, and that the section $x \in G(U)$ we get in this way satisfies $x {\mid_{U_i}} = x_i$ for each $i$. It also remains to establish the uniqueness of $x$.
In the above, you can see that our construction in the "essential surjectivity" proof amounted to specifying the object part of a quasi-inverse $\mathbf{s}$, and our construction in the "surjectivity on morphisms" proof amounted to specifying the morphism part of $\mathbf{s}$. (Note that the definition of $\mathbf{s}$ as you wrote it does not necessarily make sense if $\mathcal{B}$ is not closed under intersections.)
Best Answer
I would follow the advice of the first paragraph of my edition of Mac Lane-Moerdijk, which is to go read sections of Mac Lane's Categories for the Working Mathematician that expand on the topics treated in Chapter 1, and to solve their exercises. Other books, most notably Riehl's Category Theory in Context, are now often recommended as a first introduction to category theory. These have some improvements over Mac Lane, but it might be that the best preparation for Mac Lane-Moerdijk is Mac Lane himself. Either way should be fine. Turi's notes look nice but you want many more exercises.